# ∑f(n)=1 (n=1 to n), find f(n).

1. Jan 20, 2014

### Crack Dragon

Not sure if this is impossible, simply wondering if anti-sums are a thing, ie. (Anti)∑0.5n(n+1)=n.

2. Jan 20, 2014

### tiny-tim

Hi Crack Dragon! Welcome to PF!

Yes, it's a difference!

The difference of a series {an} is the series {an - an-1}

eg 0.5n(n+1) - 0.5(n-1)n = 0.5n((n+1) - (n-1)) = n

3. Jan 20, 2014

### Crack Dragon

Hi, thanks for that, sorry if this was in the wrong area; wasn't too sure. That's pretty handy actually, by your post count you seem to be a busy person, but do you know where I could do any reading up on summations and the like?

4. Jan 20, 2014

### tiny-tim

sorry, no idea …

but i'm sure you'll find plenty if you google it

5. Jan 21, 2014

### jsmith

Can you clarify, is the formula meant to hold for all n or for a fixed value of n? If it is for all n the f(n)=0 is the only solution in this case. If it is for a fixed n then there are infinite f(n) that will work.

6. Jan 23, 2014

### Crack Dragon

7. Jan 23, 2014

### jsmith

My mistake, I missread 1 as 0. But it doesn't change much. If the sum of f(k) from k=1 to n (btw, don't use n as both the thing you are summing over as well as the number of terms you are summing) for all n, then you must have f(1)=1 and all others are 0. To see this just set n=1 so f(1)=1 then set n=2 to get f(1)+f(2)=1 so 1+f(2)=1 which gives f(2)=0. Continue inductively to get the rest being 0.

8. Jan 23, 2014

### Jhenrique

$$\\ \sum f(n)\Delta n=1 \\ \\ \frac{\Delta\;\;}{\Delta n} \sum f(n)\Delta n = \frac{\Delta1}{\Delta n} \\ \\ f(n) = \left\{\begin{matrix} 1 & n=0\\ 0 & n\neq 0\\ \end{matrix}\right.$$

Last edited: Jan 23, 2014
9. Jan 23, 2014

### uart

The way you're using the summation symbol is wrong crack dragon. You need to sum over a "dummy" variable such as "k" in the example below.

$$\sum_{k=1}^{n} f(k) = 1$$

Now we can again ask the question: Do you require this to be valid just for some specific positive "n" or is it to be true for all positive "n"?

If it's for some specific "n" then there is a unique solution only for n=1, and an infinite number of solutions otherwise.

If it's for all positive "n" then there is just one solution, that given above by jsmith.

Last edited: Jan 23, 2014
10. Jan 23, 2014

### Staff: Mentor

What does Δ/Δn mean? I understand the operator d/dn, but have never seen similar notation using Δ.

11. Jan 23, 2014

### Jhenrique

$\Delta f=f_1-f_0$

$\frac{\Delta f}{\Delta x}=\frac{f_1-f_0}{x_1-x_0}$

How in discrete calculus we not consider an infinitesimal interval but yes a discrete interval (unitary), so Δx=1, ie:

$\frac{\Delta f}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{f(x+1)-f(x)}{1}=f(x+1)-f(x)=\Delta f$

The same logic for summation:

$\sum f(x)\Delta x=\sum f(x)\cdot 1=\sum _{x}f(x)$

FTC for the discrete case:

$\sum_{x_0}^{x_1} f(x)\Delta x=\sum_{x=x_0}^{x_1}f(x)=F(x_1+1)-F(x_0)$

inverse relationships:

$\sum \frac{\Delta f}{\Delta x}(x)\Delta x=f(x)$

$\frac{\Delta}{\Delta x} \sum f(x)\Delta x = f(x)$

with infinitesimal limits...

$\int f(x)dx=\lim_{\Delta x \rightarrow 0} \sum f(x) \Delta x$

$\frac{df}{dx}=\lim_{\Delta x \rightarrow 0} \frac{\Delta f}{\Delta x}$

computing...

$\frac{d}{dx}\left (\frac{1}{2}kx^2 \right )=kx$

$\frac{\Delta}{\Delta x}\left (\frac{1}{2}kx^2 \right )=kx+\frac{1}{2}k$

$\sum f(x)\Delta x=\frac{1}{6}kx^3-\frac{1}{4}kx^2+\frac{1}{12}kx + C$

$\int f(x)dx=\frac{1}{6}kx^3 + C$

And you still have a Z transform that is the discrete analogue of Laplace transform. All this is the discrete calculus.

Last edited: Jan 23, 2014
12. Jan 23, 2014

### Staff: Mentor

This (above) is what I've never seen before. Where does the k/2 on the right side come from?

13. Jan 23, 2014

### Jhenrique

From the definition of discrete differentiation: $\frac{\Delta f}{\Delta x} = f(x+1)-f(x)=f'(x)$

If f(x)=1/2kx² so f'(x)=kx+1/2k

Ironically or not, It's so that math works...

14. Jan 24, 2014

### Jhenrique

Now, my 2nd answer (considering the interval of summation):

The "graphic" of question below is:

$F[n]=\sum _{1}^{n}f[m]\Delta m=1$
\begin{matrix}
n & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\
F[n] & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\
F'[n] & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
\end{matrix}

Thus we have $F'[n]=\frac{\Delta F[n]}{\Delta n}=F[n+1]-F[n]=f[n]$ ([...] the colchets says that the domain is discrete). Remember that:
http://en.wikipedia.org/wiki/Kronecker_delta
http://en.wikipedia.org/wiki/Dirac_delta_function

$\delta [n]:=\left\{\begin{matrix} 1 & n=0\\ 0 & n\neq 0 \end{matrix}\right.$

So, f[n] in terms of discrete delta is:
$$f[n]= \delta [n-1]=\left\{\begin{matrix} 1 & n=1\\ 0 & n\neq 1 \end{matrix}\right.$$

I found f(n) to you. Capiche?